Daily Challenge #54 - What century is it?

Challenge
Write a function that will return an inputted numerical year in century format. The output should have the appropriate written ending ('st','nd','rd','th') as well.

Examples
In: 2259 Out: 23rd
In: 1124 Out: 12th
In: 2000 Out: 21st

Good luck!

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Comments

[E. Choroba]

This would've been even more interesting using the strict usage of "century" ;-)

#!/usr/bin/perl
use warnings;
use strict;

sub century {
    my ($year) = @_;
    my $century = 1 + int($year / 100);
    my $suffix;
    $suffix = 'th' if grep $century == $_, 11 .. 13;
    $suffix ||= {
        1 => 'st',
        2 => 'nd',
        3 => 'rd',
    }->{ substr $century, -1 } || 'th';
    $century . $suffix
}

use Test::More tests => 6;
is century(33),    '1st';
is century(2259), '23rd';
is century(1124), '12th';
is century(2000), '21st';
is century(3199), '32nd';
is century(2423), '25th';

First handle the exceptions (i.e. 11 - 13), then just use the last digit to decide.

[Olek Ria]

F#

let whatCenture (year: int) =
    let centure = year / 100 + 1

    let suffix x =
        if (centure % 13) = 12  || (centure % 13) = 11
        then "th"
        else match x % 10 with 
                | 1 -> "st"
                | 2 -> "nd"
                | 3 -> "rd"
                | _ -> "th"

    (string centure) + (suffix centure)

[SavagePixie]

Yes, yes, I know, 2000 should return "21st" century. I don't know of anyone who counts centuries like that, so my function returns them according to normal use.

const addBC = year => year < 0 ? " BC" : ""

const centurify = year => {
   const num = Math.ceil(Math.abs(year) / 100).toString()
   const suffix = num.match(/(11|12|13)$/)
      ? "th" : num.endsWith("1")
      ? "st" : num.endsWith("2")
      ? "nd" : num.endsWith("3")
      ? "rd" : "th"
   return num + suffix + addBC(year)
}

[Oleksii Filonenko]

A reasonably short and reasonably Rusty solution:

pub fn century(year: u32) -> String {
    let century = year / 100 + 1;
    let suffix = match century % 100 {
        11 | 12 | 13 => "th",
        _ => match century % 10 {
            1 => "st",
            2 => "nd",
            3 => "rd",
            _ => "th",
        },
    };
    format!("{}{}", century, suffix)
}

[Olek Ria]

Garna robota!
Good job!

[Oleksii Filonenko]

Diakuyu :)
Thanks :)

[Oleksii Filonenko]

I used your answer as a cheatsheet of sorts. Good job! :)

[SavagePixie]

Oh, good catch

[Ganesh Prasad]

A bit of functional JS

const test = require('./tester');

const century = year => {
    if (isNaN(year)) return null;
    const nYear = Number(year);
    const cent = Math.floor(nYear / 100) + 1;
    const suffix = Math.floor(cent / 10) % 10 === 1 ? 'th'
        : cent % 10 === 1 ? 'st'
        : cent % 10 === 2 ? 'nd'
        : cent % 10 === 3 ? 'rd'
        : 'th';
    return `${cent}${suffix}`;
}

test(century, [
    {
        in: [2259],
        out: '23rd',
    },
    {
        in: [1124],
        out: '12th',
    },
    {
        in: [2000],
        out: '21st'
    },
    {
        in: [11092],
        out: '111th',
    },
]);

[SavagePixie]

I like your approach, it looks very clean.

This is probably too fringe to matter in most contexts, but wouldn't your function return 111st for the year 11,092?

[Ganesh Prasad]

It would, indeed. Thanks for pointing out. Now I have fixed it and added a new test case.

OLD SOLUTION (Line 7)

const suffix = Math.floor(cent / 10) === 1 ? 'th'

UPDATED SOLUTION (Line 7)

const suffix = Math.floor(cent / 10) % 10 === 1 ? 'th'

[Jay]

Rust:

fn century_name(year: u32) -> String {
    let century = (year / 100) + 1;
    format!(
        "{}{}",
        century,
        match century % 10 {
            _ if century > 10 && century < 14 => "th", // teen numbers exception
            1 => "st",
            2 => "nd",
            3 => "rd",
            _ => "th",
        }
    )
}

[SavagePixie]

It's fixed now (I think). It should also support BC centuries.

[Olek Ria]

Thanks